Explore Measurement Concepts with Our Examples and Questions

I have somehow confused myself with this fairly straightforward proof. We need to show that $\mathrm{\lambda <!--\; \lambda \; -->}f$ is a measurable function on $(S,\mathcal{S})$, i.e. that for any $c\in <!--\; \in \; -->\mathbb{R}:\{s\in <!--\; \in \; -->S:f(s)\le <!--\; \le \; -->c\}\subset <!--\; \subset \; -->S$. If $\mathrm{\lambda <!--\; \lambda \; -->}\ne <!--\; \ne \; -->0$, then the claim follows immediately from the measurability of $f$, namely as $c/\mathrm{\lambda <!--\; \lambda \; -->}\in <!--\; \in \; -->\mathbb{R}$ it follows that $\{s\in <!--\; \in \; -->S:f(s)\le <!--\; \le \; -->c/\mathrm{\lambda <!--\; \lambda \; -->}\}=\{s\in <!--\; \in \; -->S:\mathrm{\lambda <!--\; \lambda \; -->}f(s)\le <!--\; \le \; -->c\}$
But then, if $\mathrm{\lambda <!--\; \lambda \; -->}=0,\mathrm{\lambda <!--\; \lambda \; -->}f=0$ and I am not sure how to convince myself that $\mathrm{\lambda <!--\; \lambda \; -->}f$ is measurable.

Let $T$ be a countable set, $\mathcal{A}=\{A\subseteq <!--\; \subseteq \; -->T:<!--\; :\; -->A\text{ is finite or }{A}^c\text{ is finite}\}$ and $\mathrm{\mu <!--\; \mu \; -->}:\mathcal{A}\to <!--\; \to \; -->[0,\mathrm{\infty <!--\; \infty \; -->})$ definided for every $A\in <!--\; \in \; -->\mathcal{A}$ by
$\mathrm{\mu <!--\; \mu \; -->}(A)=\{\begin{array}{ll}0,& \text{ if }A\text{ is finite }\\ 1,& \text{ if }{A}^c\text{ is finite. }\end{array}$
Then every set $A\in <!--\; \in \; -->\mathcal{A}$, such that $A^c$ is finite, is an atom with respect of $\mathrm{\mu <!--\; \mu \; -->}$.
But I can't see how it is concluded that every set $A\in <!--\; \in \; -->\mathcal{A}$, such that $A^c$ is finite, is an atom with respect of $\mathrm{\mu <!--\; \mu \; -->}$, can someone help me?

Let $f_n$ and $f$ functions of $L_1{\mathbb{R}}^n$ such that $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{f}_n=f$ almost always in ${\mathbb{R}}^n$ and ${\int <!--\; \int \; -->}_{{\mathbb{R}}^n}|f_n|dx\to <!--\; \to \; -->{\int <!--\; \int \; -->}_{{\mathbb{R}}^n}|f|dx$. Show that for every set $A\subset <!--\; \subset \; -->{\mathbb{R}}^n$ lebesgue measurable
${\int <!--\; \int \; -->}_A{f}_{n}dx\to <!--\; \to \; -->{\int <!--\; \int \; -->}_{A}fdx.$
How can I solve this problem?

My instructor briefly discussed a result in lecture that I need help with. Here was the set up.
We assumed that $\{f_n\}$ is an orthonormal system in $L^2(0,1)$ and we supposed that there exists an $M>0$ such that $|f_n(x)|\le <!--\; \le \; -->M$ a.e for all $n\in <!--\; \in \; -->\mathbb{N}$. Furthermore, we let $\{c_n\}$ be a sequence of real numbers such that $\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{c}_n{f}_n$cnfn converges a.e.
They said that it was "clear" that
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{c}_n=0.$
This is not clear or intuitive for me.
After reading through the page, I am still a bit confused. Can someone put together a working proof to clear things up?

Let $f(x,y)=\underset{n}{\sum <!--\; \sum \; -->}\frac{x}{x^2+y{n}^2}$. Show that $g(y)=\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}f(x,y)$ exists for all $y>0$. Find $g(y)$.
My first impression of this problem is to use the monotone convergence theorem (MCT) or the dominated convergence theorem (DCT) to interchange the limit and the sum. However, I do not know what to bound the function by. Thanks in advance!
If we convert the sum to an integral under the counting measure $\mathrm{\mu <!--\; \mu \; -->}$ on $\mathbb{N}$, we can re-express the function as
$f(x,y)={\int <!--\; \int \; -->}_{\mathbb{N}}\frac{x}{x^2+y{n}^2}d\mathrm{\mu <!--\; \mu \; -->}$
Ideally, we want to apply the DCT by finding an integrable function $h$ such that $|f_n|=\left|\frac{x}{x^2+y{n}^2}\right|\le <!--\; \le \; -->h$ a.e. for all $n$. But, since $h$ has to be an $L^1$ function independent of $n$ under the counting measure, DCT might not be the correct method.
Also, the DCT that I learned involves interchanging $\int <!--\; \int \; -->\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{f}_n=\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\int <!--\; \int \; -->f_n$. But, this problem statement is asking for $\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}$ and not $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}$.

Let $$ be some measurable domain and $f:E\to <!--\; \to \; -->\mathbb{R}$ a measurable map. Let $B\in <!--\; \in \; -->\mathrm{Bor}<!--\; \; -->(\mathbb{R})$ be a Borel set. Show that $f^{-<!--\; -\; -->1}(B)$ is measurable.
I'm advised to define $\mathcal{A}=\{A\mid <!--\; \mid \; -->f^{-<!--\; -\; -->1}(A)\text{ measurable}\}$. Now $\mathcal{A}$ consists of sets whose preimage is measurable, and since $f$ is continuous these sets are open. This collection seems to form a $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra on $\mathbb{R}$, but I'm confused about the construction here as it seems that $\mathcal{A}$ is the smallest $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra containing open sets, but that would mean that it's equal to $\mathrm{Bor}<!--\; \; -->(\mathbb{R})$?

I am reading one book and the author presents what he calls a distributivity formula:
$\underset{j\in <!--\; \in \; -->J}{\bigcup <!--\; \bigcup \; -->}\underset{i\in <!--\; \in \; -->I_j}{\bigcap <!--\; \bigcap \; -->}{F}_i^j=\underset{\{i_j\}\in <!--\; \in \; -->K}{\bigcap <!--\; \bigcap \; -->}\underset{j\in <!--\; \in \; -->J}{\bigcup <!--\; \bigcup \; -->}{F}_{i_j}^j$
where $K={\mathrm{\Pi <!--\; \Pi \; -->}}_{j\in <!--\; \in \; -->J}{I}_j$ (the set of all sequences {$i_j,j\in <!--\; \in \; -->J$}).
My question is regarding the notation. The LHS is quite clear to me. However, I'm now sure what are actually the members of K. Say $J=\{1,2,\dots <!--\; \dots \; -->\}$ and $I_j=\{1,2,\dots <!--\; \dots \; -->\}$. What is actually $i_j$ here?

Let $f(x)=\underset{r>0}{sup}\frac{\mathrm{\mu <!--\; \mu \; -->}(B_r(x))}{m(B_r(x))}$. Prove it is $\mathcal{B}({\mathbb{R}}^d)$ measurable. $\mathrm{\mu <!--\; \mu \; -->}$ is a measure on $\mathcal{B}({\mathbb{R}}^d)$ such that $\mathrm{\mu <!--\; \mu \; -->}(\mathcal{B}({\mathbb{R}}^d))$ is finite, and m is the Lebesgue measure.
My thoughts: suppose $x\in <!--\; \in \; -->f<a$, then there exists $\mathrm{ϵ<!--\; ϵ\; -->}$ s.t. $(1+\mathrm{ϵ<!--\; ϵ\; -->})f(x)<a$. So given some $\mathrm{\delta <!--\; \delta \; -->}>0$ for any $x^{\prime }\in <!--\; \in \; -->B_{\mathrm{\delta <!--\; \delta \; -->}}(x)$ we have
$\frac{\mathrm{\mu <!--\; \mu \; -->}(B_r(x^{\prime }))}{m(B_r(x^{\prime }))}\le <!--\; \le \; -->\frac{\mathrm{\mu <!--\; \mu \; -->}(B_{r+\mathrm{\delta <!--\; \delta \; -->}}(x))}{m(B_r(x^{\prime }))}=\frac{\mathrm{\mu <!--\; \mu \; -->}(B_{r+\mathrm{\delta <!--\; \delta \; -->}}(x))}{m(B_r(x))}=\frac{\mathrm{\mu <!--\; \mu \; -->}(B_{r+\mathrm{\delta <!--\; \delta \; -->}}(x))}{m(B_{r+\mathrm{\delta <!--\; \delta \; -->}}(x))}(\frac{r+\mathrm{\delta <!--\; \delta \; -->}}{r}{)}^d\le <!--\; \le \; -->f(x)(\frac{r+\mathrm{\delta <!--\; \delta \; -->}}{r}{)}^d$
but I need $\mathrm{\delta <!--\; \delta \; -->}$ to be uniform in order to prove $\{f<a\}$ is open and I don't know how to treat it when r is very small.

I'm trying to solve the following problem.
Let $f$ be an integrable function in (0,1). Suppose that
${\int <!--\; \int \; -->}_0^{1}fg\ge <!--\; \ge \; -->0$
for any non negative, continuous $g:(0,1)\to <!--\; \to \; -->\mathbb{R}$. Prove that $f\ge <!--\; \ge \; -->0$ a.e. in (0,1).
I'm a little unsure on what it is that I must prove in order to conclude that $f\ge <!--\; \ge \; -->0$. I tried to show that ${\int <!--\; \int \; -->}_0^1{f}^2\ge <!--\; \ge \; -->0$ but I couldn't get very far.
I'm seeking hints on how to solve this. Thanks.

I have two questions about the uniform integrability.
The definition I am using is that a class of random variables $\mathrm{\chi <!--\; \chi \; -->}$ is uniformly integrable if given an $\mathrm{ϵ<!--\; ϵ\; -->}>0$, there exists a $k$ such that for any $x$ in $\mathrm{\chi <!--\; \chi \; -->}$ we have $\mathbb{E}[|x|\mathbb{I}\{x\ge <!--\; \ge \; -->k\}]<\mathrm{ϵ<!--\; ϵ\; -->}$.
First, is that in the definition of uniform integrability, can the density function which we compute the expectation with respect to it, vary with $n$?
Second, suppose I have a sequence of random variables $(x_n{)}_{n\ge <!--\; \ge \; -->1}$ (with varying density function w.r.t n). For the two functions $f,g$, I know that $f(x)\le <!--\; \le \; -->g(x)$ for all $x$. Does the uniform integrability of the sequence $(g(x_n){)}_{n\ge <!--\; \ge \; -->1}$ imply the uniform integrability of $(f(x_n){)}_{n\ge <!--\; \ge \; -->1}$?

1.$f_n(x)$ is continuous, bounded, positive and $f_n(x)\le <!--\; \le \; -->f_{n+1}(x)$
2.$f(x)=lim{f}_n(x)$
Question: Is $f(x)$ a lower semicontinuous function?

Suppose that $f$ has a measurable domain and is continuous except at a finite number of points. Is $f$ neccessarily measurable?
Let $A$ be the set of points where $f$ is not continuous. Now since $A$ is finite we have that $m(A)=0$. Now as continuous maps are measurable if $E$ is the measurable domain, then $f$ defined on $E\setminus <!--\; \setminus \; -->A$ is measurable.
Now I have a theorem that states
For a measurable subset $D$ of $E$, f is measurable on $E$ if and only if $f{\mid <!--\; \mid \; -->}_D$ and $f{\mid <!--\; \mid \; -->}_{E\setminus <!--\; \setminus \; -->D}$ are both measurable.
$f{\mid <!--\; \mid \; -->}_{E\setminus <!--\; \setminus \; -->A}$ being measurable follows from the fact that $f$ is continuous on $E$∖$A$, but how can I show that $f$ is measurable on $A$? I only know that the measure of $A$ is zero, but nothing about the behavior of $f$ there?

Let $I_1,I_2...$ be an arrangement of the intervals $[\frac{i-<!--\; -\; -->1}{2^n},\frac{i}{2^n})$ in a sequence. If $X_n=n$ on $I_n$ and 0 elsewhere then $su{p}_{n}E{X}_n<\mathrm{\infty <!--\; \infty \; -->}$ but $P(\underset{n}{sup}{X}_n<\mathrm{\infty <!--\; \infty \; -->})=0$. My basic space is [0,1] with Lebesgue measure.
$A:=\{\underset{n}{sup}{X}_n<\mathrm{\infty <!--\; \infty \; -->}\}=\{\mathrm{\omega <!--\; \omega \; -->}\in <!--\; \in \; -->\mathrm{\Omega <!--\; \Omega \; -->}:\mathrm{\exists <!--\; \exists \; -->}M>0,\mathrm{\forall <!--\; \forall \; -->}n\in <!--\; \in \; -->\mathbb{N},X_n(\mathrm{\omega <!--\; \omega \; -->})\le <!--\; \le \; -->M\}$
hence the complement is
$\begin{array}{rl}\stackrel{¯<!--\; ¯\; -->}{A}& =\{\mathrm{\omega <!--\; \omega \; -->}\in <!--\; \in \; -->\mathrm{\Omega <!--\; \Omega \; -->}:\mathrm{\forall <!--\; \forall \; -->}M\in <!--\; \in \; -->\mathbb{N},\mathrm{\exists <!--\; \exists \; -->}n\in <!--\; \in \; -->\mathbb{N},X_n(\mathrm{\omega <!--\; \omega \; -->})>M\}\\ & =\underset{M\in <!--\; \in \; -->\mathbb{N}}{\bigcap <!--\; \bigcap \; -->}\underset{n\in <!--\; \in \; -->\mathbb{N}}{\bigcup <!--\; \bigcup \; -->}\{X_n>M\}\\ & =\underset{M\in <!--\; \in \; -->\mathbb{N}}{\bigcap <!--\; \bigcap \; -->}\underset{n\ge <!--\; \ge \; -->M}{\bigcup <!--\; \bigcup \; -->}\{X_n>M\}\end{array}$
How do i go on from here?

I need to know how precise a measurement system is, expressed as a standard deviation. I don't really have much of a background in math so please excuse me if I'm a bit unclear or inaccurate about something.
The simplest way I know of doing this is by taking e.g. 100 measurements of the same object and then using the usual sample standard deviation formula.
Unfortunately, I can't use this method. I must measure different objects multiple times. Let's say 10 objects, 10 times each.
The thing that comes to mind is to calculate the mean for each group, subtract the relevant means from each measurement (sort of a normalisation) and then calculate the standard deviation from the resulting 100 numbers. Does this make sense? If not, how do I approach the problem?

In a study looking at undergraduate students’ perceptions of sense of community at their university, a researcher hypothesizes that the farther away students live from campus (in miles), the less they feel they are part of the university community. The researcher collected data for the following two variables – miles from campus and part of community (rating from 1-10 of how much they felt part of the university community).
what is the scale of measurement? (i.e., nominal, ordinal, interval or ratio)

Let $M,N$ be smooth manifolds, and let $F:<!--\; :\; -->M\to <!--\; \to \; -->N$ be a smooth immersion. I know that $F$ is a local embedding, but is it also an embedding almost everywhere?
In other words, does there exists a set of measure zero $X\subset <!--\; \subset \; -->M$ such that $F$ restricted to $M\setminus <!--\; \setminus \; -->X$ is a smooth embedding?

Let $w:\mathbb{N}\to <!--\; \to \; -->[0,\mathrm{\infty <!--\; \infty \; -->})$ continuous.
For each $f:\mathbb{N}\to <!--\; \to \; -->\mathbb{C}$ such that ${\displaystyle \underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}w(n)f(n)}$ is absolutely convergent we define $\mathrm{\Lambda <!--\; \Lambda \; -->}f={\displaystyle \underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}w(n)f(n)}$
It is easy to prove that $\mathrm{\Lambda <!--\; \Lambda \; -->}$ is linear and satisfies that $\mathrm{\forall <!--\; \forall \; -->}f:f(\mathbb{N})\subset <!--\; \subset \; -->[0,\mathrm{\infty <!--\; \infty \; -->})\Rightarrow <!--\; \Rightarrow \; -->\mathrm{\Lambda <!--\; \Lambda \; -->}f\in <!--\; \in \; -->[0,\mathrm{\infty <!--\; \infty \; -->})$
(I think this last property has a name but I don't know what it is)
By the Riesz representation theorem there is only one positive measure $\mathrm{\nu <!--\; \nu \; -->}$ such that $\mathrm{\Lambda <!--\; \Lambda \; -->}f={\displaystyle {\int <!--\; \int \; -->}_{\mathbb{N}}fd\mathrm{\nu <!--\; \nu \; -->}}$. How can I find the measure $\mathrm{\nu <!--\; \nu \; -->}$ that fulfills this property?

Let $(X,\mathcal{M},\mathrm{\mu <!--\; \mu \; -->})$ be a measueable space, and let $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ be nonnegative $f(x)\ge <!--\; \ge \; -->0$ and measurable. Let $E_n=\{x\in <!--\; \in \; -->\mathbb{R}:f(x)\ge <!--\; \ge \; -->\frac{1}{n}\}$. Show the following result
${\int <!--\; \int \; -->}_{\mathbb{R}}fdm=\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\int <!--\; \int \; -->}_{[-<!--\; -\; -->n,n]}fdm=\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\int <!--\; \int \; -->}_{E_n}fdm.$
I was thinking maybe we can assume $f{\mathrm{\chi <!--\; \chi \; -->}}_{E_n}\le <!--\; \le \; -->f$, monotone increasing sequence of nonnegative measurable functions, and then apply monotone convergence theorem, but I am not sure that we can find such increasing sequence, as $\frac{1}{n}$ is decreasing.

let $f_n:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ be a nonnegative sequence of Borel measurable functions which converge to a function $f$, and the integral of all $f_n$ and of $f$ are all 1. Then does the integral of $|f_n-<!--\; -\; -->f|$ goes to 0 as $n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}?$?

Let $X=Y$ be uncountable and define $\mathcal{A},\mathcal{B}$ to be the countable-cocountable $\mathrm{\sigma <!--\; \sigma \; -->}$-algebras on $X,Y$ respectively. Let $C=\mathrm{\sigma <!--\; \sigma \; -->}(A\times <!--\; \times \; -->B)$. Prove that for each subset $E\in <!--\; \in \; -->C$, there exist $A,B$ in $\mathcal{A},\mathcal{B}$ countable such that either $E\subseteq <!--\; \subseteq \; -->(A\times <!--\; \times \; -->Y)\cup <!--\; \cup \; -->(X\times <!--\; \times \; -->B)$ or $E^c\subseteq <!--\; \subseteq \; -->(A\times <!--\; \times \; -->Y)\cup <!--\; \cup \; -->(X\times <!--\; \times \; -->B)$.
My idea is to define $D$ as the collection of all such subsets of $C$, prove that $D$ contains $A_1\times <!--\; \times \; -->B1$ for each $A_1\in <!--\; \in \; -->\mathcal{A},B_1\in <!--\; \in \; -->\mathcal{B}$ and then show that $D$ is a $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra.
I am able to prove that $D$ is a $\mathrm{\sigma <!--\; \sigma \; -->}$ algebra, and I can show that in the case $A_1,B_1$ are either both countable or cocountable, $A_1\times <!--\; \times \; -->B_1$ is in $D$. How do I do it for the case that exactly one of them is cocountable?
Something that may be useful is
$(A_1\times <!--\; \times \; -->B_1{)}^c=(A_1^c\times <!--\; \times \; -->Y)\cup <!--\; \cup \; -->(X\times <!--\; \times \; -->B_1^c)\cup <!--\; \cup \; -->(A_1^c\times <!--\; \times \; -->B_1^c)$